Asked by kruthi
the fourth term of an arithmetic progression is equal to 3 times the first term and the seventh term exceeds twice the third term by 1 find the first term and the common differrence
Answers
Answered by
Steve
a+3d = 3a
a+6d = 2(a+2d)+1
a=3
d=2
chec:
3+6 = 3*3
3+12 = 2(7)+1
a+6d = 2(a+2d)+1
a=3
d=2
chec:
3+6 = 3*3
3+12 = 2(7)+1
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